Theo bài ra ta có:
\(\dfrac{a}{b}=\dfrac{3}{7}\Rightarrow7a=3b\Rightarrow a=\dfrac{3b}{7}\)
Thay vào \(ab=525\) ta có:
\(\dfrac{3b}{7}\cdot b=525\Rightarrow3b^2=3675\)
\(\Rightarrow b^2=1225\Rightarrow b=\pm35\)
*)Xét \(b=35\Rightarrow a=\dfrac{3b}{7}=\dfrac{3\cdot35}{7}=15\)
*)Xét \(b=-35\Rightarrow a=\dfrac{3b}{7}=\dfrac{3\cdot\left(-35\right)}{7}=-15\)
Cách 2:
Giải:
Ta có: \(\dfrac{a}{3}=\dfrac{b}{7}\)
Đặt \(\dfrac{a}{3}=\dfrac{b}{7}=k\Rightarrow\left\{{}\begin{matrix}a=3k\\b=7k\end{matrix}\right.\)
Mà \(a.b=525\)
\(\Rightarrow3.k.7.k=525\)
\(\Rightarrow21.k^2=525\)
\(\Rightarrow k^2=25\)
\(\Rightarrow k=\pm5\)
+) \(k=5\Rightarrow x=15,y=35\)
+) \(k=-5\Rightarrow x=-15,y=-35\)
Vậy cặp số \(\left(x;y\right)\) là \(\left(15;35\right);\left(-15;-35\right)\)
